A) 73
B) 77
C) 83
D) 87
Correct Answer: A
Solution :
\[PC{{l}_{5}}PC{{l}_{3}}+C{{l}_{2}}\] Normal molecular weight of\[PC{{l}_{5}}\] \[=2\times \] vapour density \[=2\times 60=120\] Observed molecular weight of \[PC{{l}_{5}}=30+5\times 35.5\] \[=207.5\] \[i=\frac{Observed\text{ }molecular\text{ }weight}{Normal\text{ }molecular\text{ }weight}=\frac{207.5}{120}=1.73\] Degree of dissociation \[=\frac{i-1}{n-1}=\frac{1.73-1}{2-1}=0.73\] % dissociation\[=0.73\times 100=73%.\]
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